The half-life of ${}_6{C}^{14}$ if its K or \lambda is $2.31\times {10}^{-4}$ is

The correct option is B $3\times {10}^{3}yrs$
$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{2.31\times {10}^{-4}}=0.3\times {10}^{4}yrs$
$=3.0\times {10}^{3}yrs$.